// 原数据结构
const list = [
  {
    "keyIndex": 0,
    "id": 115,
    "skuId": "1582625146908418050",
    "skuCode": "DTDLSPUHSM00ALZ9H",
    "a29697409026": "1582289829831626755",
    "a74899724290": "1567482277897674753",
    "a11715188737": "1582622511887155202",
    "a41246175233": "1582679152510853122"
  },
  {
    "keyIndex": 1,
    "id": 116,
    "skuId": "1582625147151687681",
    "skuCode": "DTDLSPUBSM005JIE2",
    "a29697409026": "1582289829831626755",
    "a74899724290": "1581963282939617282",
    "a11715188737": "1582622511874572289",
    "a41246175233": "1582623041812406273"
  },
  {
    "keyIndex": 2,
    "id": 189,
    "skuId": "1583003724091211778",
    "skuCode": "DTDLSPUBSNSXY95Z0",
    "a29697409026": "1582289829831626755",
    "a74899724290": "1581963282939617282",
    "a11715188737": "1582622511874572289",
    "a41246175233": "1582623041812406273"
  }
]

// 在不关心id等其他参数相不相同的情况下，考虑a29697409026、a74899724290、a11715188737、a41246175233四项对应的value值全等的话为重复数据，则去重

// 这边通过Map数据结构处理，当然也有好几种循环可以实现：
// 把四项的value值当做map的key值储存，然后判断是否存在这个key值，存在则不保存，反之：

let m = new Map();
for (let item of list) {
  if (!m.has(item['a29697409026'] + item['a74899724290'] + item['a11715188737'] + item['a41246175233'])) {
    m.set(item['a29697409026'] + item['a74899724290'] + item['a11715188737'] + item['a41246175233'], item);
  }
}
arr = [...m.values()];
console.log(arr)

// result: [
//   {
//     "keyIndex": 0,
//     "id": 115,
//     "skuId": "1582625146908418050",
//     "skuCode": "DTDLSPUHSM00ALZ9H",
//     "a29697409026": "1582289829831626755",
//     "a74899724290": "1567482277897674753",
//     "a11715188737": "1582622511887155202",
//     "a41246175233": "1582679152510853122"
//   },
//   {
//     "keyIndex": 1,
//     "id": 116,
//     "skuId": "1582625147151687681",
//     "skuCode": "DTDLSPUBSM005JIE2",
//     "a29697409026": "1582289829831626755",
//     "a74899724290": "1581963282939617282",
//     "a11715188737": "1582622511874572289",
//     "a41246175233": "1582623041812406273"
//   },
// ]
